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For a sphere of radius $a$ rotating with angular velocity $\Omega$, the primary Stokes flow has $v_{\varphi }=K\sin \theta /r^2$, $K=\Omega a^3$. In the weak, inertialess second-order fluid correction, put $c={\psi }_1+2{\psi }_2$. The stream function in the title hasBoth meridional components vanish at $r=a$ and decay at infinity. Indeed, $E_{r\varphi }=-3K\sin \theta /(2r^3)$ gives $[\nabla \times \nabla \cdot (2c{E}^2){]}_{\varphi }=72c{K}^2\sin \theta \cos \theta /r^8$, and the curl of the momentum equation reduces to $(d^2/d{r}^2-6/r^2{)}^{2}H=72c{K}^2/(\mu r^7)$. For $c>0$ the secondary flow enters towards the equator and leaves near the poles; $c<0$ reverses it and $c=0$ removes this correction. The meridional circulation is even in the sign of sphere rotation. Inertial secondary flow is excluded from this calculation.

Linearizing the differential pom-pom model about its relaxed state leaves no first-order stretch perturbation in incompressible flow. Its anisotropic orientation obeys $\dot{a}+a/{\tau }_1=A_1$, producing the displayed relaxation modulus. In a slow-flow expansion, $A=I+{\tau }_1{A}_1+{\tau }_1^2(2A_1^2-A_2)+O(3)$. Trace normalization and the first stretch correction affect only isotropic stress at this order, giving the displayed second-order fluid coefficients. Linear small-amplitude response allows arbitrary frequency, whereas this local second-order expansion also requires slow time dependence.

The usual elastic secondary flow around a rotating sphere is inward towards the equator, then poleward beside the sphere and outward near both poles. It is superposed on the primary azimuthal rotating sphere in Stokes flow. Tensile normal stresses along curved azimuthal streamlines act as hoop stresses, pulling liquid towards the rotation axis. Incompressibility supplies the axial return flow. Unlike the primary rotation, this meridional circulation does not reverse when the rotation is reversed.

The direction assumes the usual polymeric sign $c={\psi }_1+2{\psi }_2>0$ and negligible inertia. A general second-order fluid has circulation proportional to $c$: a negative value reverses the arrows and $c=0$ eliminates the leading elastic circulation. Inertial secondary circulation is a different contribution, so slow rotation alone does not determine which contribution dominates. The sketch shows the elastic contribution for $c>0$:

For a quantitative check of the sketch, take sphere radius $a$, angular velocity $\Omega$ and $K=\Omega a^3$. The primary azimuthal velocity is $v_{\varphi }=K\sin \theta /r^2$, with $\theta$ measured from the rotation axis. Its only nonzero strain components in the spherical orthonormal basis are $E_{r\varphi }=E_{\varphi r}=-3K\sin \theta /(2r^3)$. The leading elastic secondary circulation around a rotating sphere has the stream functionwith $v_r=H(3{\cos }^2\theta -1)/r^2$ and $v_{\theta }=-H^{\prime }\sin \theta \cos \theta /r$. The bracket is $(1-a/r{)}^2(1+2a/r)>0$ outside the sphere. Thus $v_r<0$ at the equator and $v_r>0$ near the poles when $c>0$, and both components vanish on the sphere. The diagram uses these velocities, not independently chosen arrows.

Two other manifestations of the same elastic normal-stress differences are rod climbing, in which inward hoop stresses raise a free surface around a rotating rod, and die swell, in which relaxation of the axial stress and stored polymer deformation allows an extrudate to expand sideways after leaving the die. These are not explained by a scalar shear viscosity alone. A Newtonian extrudate can also swell through velocity-profile rearrangement, so the elastic contribution should not be identified with every instance of swelling.

The second-order fluid is a slow, weak-flow expansion of the objective constitutive history of an isotropic simple fluid. Material frame indifference requires objective tensor rates. For an incompressible flow, the leading isotropic stress is arbitrary pressure, the linear deviatoric term is $2\mu E$, and at the next order the independent contributions can be represented by $E^2$ and an objective rate of $E$. Isotropic scalar contributions are absorbed into the pressure. Their coefficients can be expressed through the two low-rate normal-stress differences, givingHere $\stackrel{△}{E}$ is the lower-convected derivative, not a division by $E$. This local expansion is appropriate when deformation and temporal change over the relevant viscoelastic relaxation time are small: both the Weissenberg number and the appropriate Deborah number must be small. It captures leading normal stresses in slow flows, but not arbitrary finite-strain history, rapid transients or strongly nonlinear shear thinning. In particular its restrictions are stronger than small-amplitude linear oscillatory response at arbitrary frequency.

For steady simple shear flow $\mathbf{v}=(\dot{\gamma }y,0,0)$, use component-first $L_{ij}={\partial }_j{v}_i$, so the paper's $\nabla v$ is $L^T$. The relevant tensors areSubstitution givesThus the first and second normal-stress differences are $N_1={\psi }_1{\dot{\gamma }}^2$ and $N_2={\psi }_2{\dot{\gamma }}^2$, and the shear stress is $\mu \dot{\gamma }$.

For the velocity-preservation results, the definitions imply $\stackrel{△}{E}=\stackrel{\circ }{E}+2E^2$. Consequently the non-Newtonian stress can be writtenwhere $\stackrel{\circ }{E}$ is the Jaumann derivative. Let ${\mathbf{v}}_N$ solve the steady incompressible Stokes flow problem with the prescribed boundary velocities. The supplied identity says that $\nabla \cdot {\stackrel{\circ }{E}}_N$ is curl-free. If ${\psi }_1+2{\psi }_2=0$, then $\nabla \cdot S_N=\nabla \Phi$ for a scalar $\Phi$ on the ordinary simply connected flow domain. The momentum equation at ${\mathbf{v}}_N$ isHence the Newtonian velocity itself is a solution of the second-order constitutive model, with altered pressure $p=p_N+\Phi$. For nontrivial topology one also needs the curl-free force to admit a single-valued pressure; this is implicit in the usual boundary-value problem.

On the regular weakly non-Newtonian branch, Uniqueness of Stokes flow fixes the velocity correction. At first order it solves homogeneous Stokes flow, with zero boundary velocity, after the gradient force has been absorbed into pressure. Multiplying its equation by the correction and integrating gives $2\mu \int E_1:E_{1}dV=0$, hence the correction vanishes. Higher perturbative corrections vanish in turn because the previous velocities remain ${\mathbf{v}}_N$. The velocity is $\mathbf{v}={\mathbf{v}}_N$ in this second-order-fluid solution; only the pressure and stresses change. This is Newtonian velocity preservation in a second-order fluid. It does not assert global uniqueness of distant nonlinear solution branches or exact equality for a real fluid whose higher-order constitutive terms have been omitted.

For a planar incompressible velocity $(u(x,y),v(x,y),0)$, the rate-of-strain tensor has the formAll quantities are independent of $z$, so $\nabla \cdot E_N^2=\nabla (a^2+b^2)$. The extra quadratic term is therefore also a pressure gradient. Together with the supplied identity for the Jaumann derivative, the entire non-Newtonian force is a gradient for arbitrary ${\psi }_1,{\psi }_2$. The same boundary-value and weak-branch argument proves $\mathbf{v}={\mathbf{v}}_N$ for every such planar second-order-fluid flow with prescribed boundary velocities, without requiring ${\psi }_1+2{\psi }_2=0$. Specifying stresses or a free surface instead of velocities would not give the same conclusion automatically, because the extra stresses affect those boundary conditions.

An isotropic simple fluid has local stress determined by the deformation history of the material element. A viscometric flow is one whose relative deformation history is that of steady simple shear flow, up to rigid changes of frame. At a given element the shear rate is constant along its history, although it can differ between elements. This includes the locally simple-shear kinematics used in standard viscometric measurements.

Use the component-first velocity gradient $K_{ij}=\partial v_i/\partial x_j$. The first two Rivlin-Ericksen tensors areIn the axes of a steady shear $v_x=sy$, the relative deformation gradient over a lag $q$ is $F=I+qK$, since $K^2=0$. The material history is determined by $s$, and its tensor symmetries exclude $xz$ and $yz$ stress components. The remaining symmetric stress consists of an isotropic part, one shear component, and two independent normal-stress differences. The relevant matrices areThus $I,A_1,A_1^2,A_2$ span the allowed stress components. Material isotropy and reversal of the shear show that their scalar coefficients depend on $|s|=(A_1:A_1/2{)}^{1/2}$. Define the viscometric functions bySolving for the matrix coefficients givesThis is an exact steady-viscometric representation, rather than an assumption that the fluid is everywhere a second-order fluid. The shear viscosity $\mu$ measures the shear resistance, while ${\psi }_1$ and ${\psi }_2$ measure the first and second normal-stress differences generated by shear. Their zero-rate values are understood as limits when these exist. An isotropic pressure changes neither normal difference.

For a flow in the lubrication approximation with gap $H$, longitudinal scale $\ell \gg H$ and speed $U$, the shear-rate scale is $U/H$. The viscous shear force per unit volume scales as $\mu U/H^2$. Quadratic viscoelastic normal stresses scale as ${\psi }_i{U}^2/H^2$; their longitudinal gradients, and the comparable mixed-component corrections, contribute forces of scale ${\psi }_i{U}^2/(H^2\ell )$. Therefore the normal-stress smallness in lubrication criterion isEquivalently, the thin-gap aspect ratio must multiply the shear-based elastic parameter to give a small quantity: $(H/\ell )|{\psi }_i|\dot{\gamma }/\mu \ll 1$. Normal stresses need not themselves be smaller than the shear stress; it is their effect on the leading force balance that matters. If the flow changes on an additional short time scale $T$, the $D_t{A}_1$ term also requires $|{\psi }_1|/(\mu T)\ll 1$ for its neglect. Material coefficients should be evaluated at the characteristic shear rate.

Now take positive cylinder rotation, measure $\theta$ counterclockwise from the rightward horizontal, and let $y=r-a$ increase outward through the film. The tangent has upward component $\cos \theta$, so the tangential gravitational acceleration is $-g\cos \theta$. To leading order in $h/a$, neglect inertia and higher-order hydrostatic or capillary pressure gradients, and write $U=\Omega a$. Tangential force balance and the shear-free free surface givehenceThe wall shear stress is $-\rho gh\cos \theta$. It opposes rotation on the uphill side and assists it on the downhill side.

For a power-law fluid, ${\sigma }_{\theta y}=k|u_y{|}^{n-1}{u}_y$. Its inversion must retain the sign:Integrating with the no-slip boundary condition $u(0)=U$ yieldsThe volume flux per unit axial length is $Q={\int }_0^{h}u(y)dy$. The integrated bracket is $[(n+1)/(2n+1)]h^{(2n+1)/n}$, so the power-law coating on a rotating cylinder hasAt $\cos \theta =0$ the gravitational term vanishes by continuity. Steady mass conservation makes $Q$ independent of $\theta$.

On the uphill half, put $D=(\rho g\cos \theta /k{)}^{1/n}$ and $m=(2n+1)/n$. The local flux is $Q(h)=Uh-(D/m)h^m$, withIt has a unique maximum atA positive coating must carry the same steady flux at every angle. The tightest local bound occurs at $\cos \theta =1$, not on the downhill side, where the flux increases monotonically with thickness. ConsequentlyFor $0<Q<Q_{\max }$ the thinner positive root on the uphill side connects through $\cos \theta =0$ to the unique downhill root, giving a periodic leading-order coating profile. At the limiting flux the two uphill roots meet at $\theta =0$; it is the maximum flux of this lubrication model, with a possible limiting corner resolved by neglected higher-order effects. For $n=1$ the check is $Q_{\max }=(2/3)U\sqrt{kU/(\rho g)}$. All these conclusions require the resulting film to remain thin enough for the lubrication approximation and for the neglected normal stresses and inertia to satisfy their scaling restrictions.

At rest the tensor equation relaxes $A$ toward $I$ and the stretch equation relaxes $\lambda$ toward one. ThusThe equilibrium nonpressure stress $GI/3$ is isotropic and can be absorbed into the arbitrary pressure in incompressible flow.

For the linear response, put $A=I+a+O({\epsilon }^2)$, $\lambda =1+\ell +O({\epsilon }^2)$ and take the velocity gradient to be $O(\epsilon )$. The first-order tensor equation isIts matrix trace has no forcing, because incompressibility gives $\mathrm{tr}K=0$. Starting relaxed, $\mathrm{tr}a=0$. The first-order stretch equation is $D_t\ell +\ell /{\tau }_2=(I/3):K=0$, hence $\ell =0$. Consequently $B=I/3+a/3+O({\epsilon }^2)$, and its anisotropic stress is $Ga/3$. From the relaxed remote past,The linear and second-order response of the differential pom-pom model therefore hasThe stretch time ${\tau }_2$ does not enter the linear shear relaxation modulus.

For the second-order fluid constants, additionally make the slow-flow expansion: count $K$ as first order and $D_t$ as raising the order by one. IteratinggivesUsing the Rivlin-Ericksen tensor convention $A_2=D_t{A}_1+A_{1}K+K^T{A}_1$, the identityreduces this to $A=I+{\tau }_1{A}_1+{\tau }_1^2(2A_1^2-A_2)+O(3)$. Since $\mathrm{tr}{A}_2=\mathrm{tr}{A}_1^2$, normalization givesStretch first changes at second order. Its equation gives $\lambda -1={\tau }_1{\tau }_2\mathrm{tr}(A_1^2)/6+O(3)$. Multiplication by ${\lambda }^2$ therefore adds only an isotropic stress at second order; multiplying its second-order increment by the anisotropic first-order term would be third order. Absorb these isotropic corrections into $p_{\ast }$ to obtainComparison with the viscometric functions parametrization yieldsEquivalently, in $\sigma =-pI+{\mu }_0{A}_1+{\alpha }_1{A}_2+{\alpha }_2{A}_1^2$, the constants are ${\alpha }_1=-G{\tau }_1^2/3$ and ${\alpha }_2=2G{\tau }_1^2/3$. The general small-amplitude linear response above permits fast oscillation; the local second-order fluid expansion also requires the slow-flow ordering just used.

The local motion is simple shear with flow direction $z$, gradient direction $r$, and vorticity direction $\theta$. Writing $q=du/dr$, a second-order fluid has first and second normal-stress differencesUp to an isotropic contribution absorbed into pressure, a convenient representation isThus both the flow-direction and gradient-direction normal stresses can be nonzero, while the radial momentum equation becomesThe first normal-stress coefficient affects ${\tau }_{zz}$ but cancels from radial balance.

In the yielded layer,Moreover,Consequently $dp/dr$ has the sign opposite to ${\Psi }_2$ and is independent of ${\Psi }_1$. For the common case ${\Psi }_2<0$, pressure increases radially outward toward the free surface.

Using the component-first velocity gradient $K_{ij}=\partial v_i/\partial x_j$, the Rivlin-Ericksen tensors are defined by the displayed recurrence. In particular $A_1$ is twice the rate-of-strain tensor. These tensors transform objectively under time-dependent rigid observer changes. With this convention the upper-convected derivative of $A_1$ is $A_2-2A_1^2$, an identity useful in the slow-flow expansion of a second-order fluid.